problem 11

Internal problem ID [10567]
Internal ﬁle name [OUTPUT/9515_Monday_June_06_2022_03_02_52_PM_27537007/index.tex]

Book: Handbook of exact solutions for ordinary diﬀerential equations. By Polyanin and Zaitsev. Second edition
Section: Chapter 1, section 1.2. Riccati Equation. subsection 1.2.7-2. Equations containing arccosine.
Problem number: 11.
ODE order: 1.
ODE degree: 1.

\[ \boxed {y^{\prime }-y^{2}-\lambda x \arccos \left (x \right )^{n} y=\arccos \left (x \right )^{n} \lambda } \]

15.2.1 Solving as riccati ode

In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= y^{2}+\arccos \left (x \right )^{n} \lambda x y +\arccos \left (x \right )^{n} \lambda \end {align*}

This is a Riccati ODE. Comparing the ODE to solve \[ y' = y^{2}+\arccos \left (x \right )^{n} \lambda x y +\arccos \left (x \right )^{n} \lambda \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\arccos \left (x \right )^{n} \lambda \), \(f_1(x)=\arccos \left (x \right )^{n} \lambda x\) and \(f_2(x)=1\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{u} \tag {1} \end {align*}

Using the above substitution in the given ODE results (after some simpliﬁcation)in a second order ODE to solve for \(u(x)\) which is \begin {align*} f_2 u''(x) -\left ( f_2' + f_1 f_2 \right ) u'(x) + f_2^2 f_0 u(x) &= 0 \tag {2} \end {align*}

But \begin {align*} f_2' &=0\\ f_1 f_2 &=\arccos \left (x \right )^{n} \lambda x\\ f_2^2 f_0 &=\arccos \left (x \right )^{n} \lambda \end {align*}

Substituting the above terms back in equation (2) gives \begin {align*} u^{\prime \prime }\left (x \right )-\arccos \left (x \right )^{n} \lambda x u^{\prime }\left (x \right )+\arccos \left (x \right )^{n} \lambda u \left (x \right ) &=0 \end {align*}

\[ u \left (x \right ) = x \left (\left (\int {\mathrm e}^{\int \frac {\arccos \left (x \right )^{n} \lambda \,x^{2}-2}{x}d x}d x \right ) c_{1} +c_{2} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \left (\int {\mathrm e}^{\int \frac {\arccos \left (x \right )^{n} \lambda \,x^{2}-2}{x}d x}d x \right ) c_{1} +c_{2} +x \,{\mathrm e}^{\int \frac {\arccos \left (x \right )^{n} \lambda \,x^{2}-2}{x}d x} c_{1} \] Using the above in (1) gives the solution \[ y = -\frac {\left (\int {\mathrm e}^{\int \frac {\arccos \left (x \right )^{n} \lambda \,x^{2}-2}{x}d x}d x \right ) c_{1} +c_{2} +x \,{\mathrm e}^{\int \frac {\arccos \left (x \right )^{n} \lambda \,x^{2}-2}{x}d x} c_{1}}{x \left (\left (\int {\mathrm e}^{\int \frac {\arccos \left (x \right )^{n} \lambda \,x^{2}-2}{x}d x}d x \right ) c_{1} +c_{2} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution

\[ y = \frac {-\left (\int {\mathrm e}^{\int \frac {\arccos \left (x \right )^{n} \lambda \,x^{2}-2}{x}d x}d x \right ) c_{3} -1-x \,{\mathrm e}^{\int \frac {\arccos \left (x \right )^{n} \lambda \,x^{2}-2}{x}d x} c_{3}}{x \left (\left (\int {\mathrm e}^{\int \frac {\arccos \left (x \right )^{n} \lambda \,x^{2}-2}{x}d x}d x \right ) c_{3} +1\right )} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {-\left (\int {\mathrm e}^{\int \frac {\arccos \left (x \right )^{n} \lambda \,x^{2}-2}{x}d x}d x \right ) c_{3} -1-x \,{\mathrm e}^{\int \frac {\arccos \left (x \right )^{n} \lambda \,x^{2}-2}{x}d x} c_{3}}{x \left (\left (\int {\mathrm e}^{\int \frac {\arccos \left (x \right )^{n} \lambda \,x^{2}-2}{x}d x}d x \right ) c_{3} +1\right )} \\ \end{align*}

Veriﬁcation of solutions

15.2.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-y^{2}-\lambda x \arccos \left (x \right )^{n} y=\arccos \left (x \right )^{n} \lambda \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{2}+\lambda x \arccos \left (x \right )^{n} y+\arccos \left (x \right )^{n} \lambda \end {array} \]

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
trying Bernoulli 
trying separable 
trying inverse linear 
trying homogeneous types: 
trying Chini 
differential order: 1; looking for linear symmetries 
trying exact 
Looking for potential symmetries 
found: 2 potential symmetries. Proceeding with integration step`

\[ y \left (x \right ) = \frac {{\mathrm e}^{\int \frac {\arccos \left (x \right )^{n} \lambda \,x^{2}-2}{x}d x} x +\int {\mathrm e}^{\int \frac {\arccos \left (x \right )^{n} \lambda \,x^{2}-2}{x}d x}d x -c_{1}}{\left (c_{1} -\left (\int {\mathrm e}^{\int \frac {\arccos \left (x \right )^{n} \lambda \,x^{2}-2}{x}d x}d x \right )\right ) x} \]

\begin{align*} y(x)\to -\frac {x \int _1^x\frac {\exp \left (2^{-n-3} \lambda \arccos (K[1])^n \left (\arccos (K[1])^2\right )^{-n} \left (\Gamma (n+1,2 i \arccos (K[1])) (-i \arccos (K[1]))^n+(i \arccos (K[1]))^n \Gamma (n+1,-2 i \arccos (K[1]))\right )\right )}{K[1]^2}dK[1]+\exp \left (\lambda 2^{-n-3} \arccos (x)^n \left (\arccos (x)^2\right )^{-n} \left ((-i \arccos (x))^n \Gamma (n+1,2 i \arccos (x))+(i \arccos (x))^n \Gamma (n+1,-2 i \arccos (x))\right )\right )+c_1 x}{x^2 \left (\int _1^x\frac {\exp \left (2^{-n-3} \lambda \arccos (K[1])^n \left (\arccos (K[1])^2\right )^{-n} \left (\Gamma (n+1,2 i \arccos (K[1])) (-i \arccos (K[1]))^n+(i \arccos (K[1]))^n \Gamma (n+1,-2 i \arccos (K[1]))\right )\right )}{K[1]^2}dK[1]+c_1\right )} \\ y(x)\to -\frac {1}{x} \\ \end{align*}

15.2 problem 11

15.2.1 Solving as riccati ode

15.2.2 Maple step by step solution